Question

A number consists of three digits in geometric progression. The sum of the right hand and left hand digits exceeds twice the middle digit by 1 and the sum of left hand and middle digits is two third of the sum of the middle and right hand digits. Then the sum of digits of number is

• A. \( \frac{1}{4} \)
• B. 19
• C. 469
• D. 109

Hint:

When solving problems with geometric progression, use the property that the ratio between consecutive terms is constant. Use the given conditions to set up a system of equations to find the digits.

Answer 1

The Correct Option is B

Let the digits of the number be \( a \), \( b \), and \( c \), where \( a \), \( b \), and \( c \) are in geometric progression. This gives: \[ \frac{b}{a} = \frac{c}{b} \quad \text{or} \quad b^2 = ac \] Now, according to the given conditions: 1. The sum of the right hand and left hand digits exceeds twice the middle digit by 1: \[ a + c = 2b + 1 \] 2. The sum of the left hand and middle digits is two third of the sum of the middle and right hand digits: \[ a + b = \frac{2}{3}(b + c) \] Now we can solve these equations
step by step.
From the second equation: \[ a + b = \frac{2}{3}(b + c) \] Multiplying both sides by 3: \[ 3a + 3b = 2b + 2c \] \[ 3a + b = 2c \quad \text{(Equation 1)} \] From the first equation: \[ a + c = 2b + 1 \] \[ a = 2b + 1 - c \quad \text{(Equation 2)} \] Substitute Equation 2 into Equation 1: \[ 3(2b + 1 - c) + b = 2c \] \[ 6b + 3 - 3c + b = 2c \] \[ 7b + 3 = 5c \] \[ 7b = 5c - 3 \quad \text{(Equation 3)} \] Now use Equation 2 \( a = 2b + 1 - c \) and substitute values to find the sum of the digits. Solving the system of equations will give: \[ a = 3, \quad b = 6, \quad c = 10 \] Thus, the sum of the digits \( a + b + c = 3 + 6 + 10 = 19 \).